If k \lt (\sqrt{2}+1)^{3} \lt k+2, where k is a natural number, then what is the value of k?
- A. 11
- B. 13 ✓
- C. 15
- D. 17
Correct Answer: B. 13
Explanation
Expanding (\sqrt{2}+1)^3 = 2\sqrt{2} + 6 + 3\sqrt{2} + 1 = 7 + 5\sqrt{2}. Using \sqrt{2} \approx 1.414, the value is approximately 7 + 5(1.414) = 14.07. We need k \lt 14.07 \lt k+2. Testing the options, k=13 gives 13 \lt 14.07 \lt 15, which is true.
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